By samdaman on December 2nd, 2010

Differentiation: Product Rule

Differentiation is a part of calculus when you find the derivative of a function, however in some equations you are asked or need to find the derivative of the product of two or more functions.

First and foremost lets start with the basic formula for the product rule.

(f.g)’ = f’.g + f.g’

or with the Leibniz notation (this may be a bit tricky to see) f can be subbed for u in this and g for v for simplicity sake of both formula using the same letters. Remember to substitute the u and v in dv and du as well.

(d/dx)(u.v) = (u.(dv/dx)) + (v.(du/dx))

I prefer the first formula as it’s obviously more easy to see and grasp.

So as we already know how to differentiate or I imagine you wouldn’t be finding yourself here reading this article I can assume you know what the above symbols mean, just a quick reminder if you are a real starter to this and need a quick heads up.

What both formula are showing is the derivative of the products f and g (left hand side of the equation shown by the (f.g)’ or ((d/dx)(u.v)), the d/dx has a longer explanation which is confusing and time consuming so we’ll leave it as the way the lazy mathematicians symbolize taking the derivative of a function (products of the functions in the case of this article) so the d/dx in a way acts like the apostrophe outside of the brackets in the first formula enclosing the (f.g) it is a way of symbolising the mathematician to take the derivative of the function) Get it now? If not have a read back as it does make sense it may just be a bit tricky to get your head around.

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The right side shows (for simplicity again i’ll use the top formula) (derivative of f times g) + (f times the derivative of g) simple right.

So lets put it into action. I’ll put some questions down and you try them, afterwards scroll down and I’ll work through them with you. If you don’t know what the following symbols mean I’ll jot down a quick key.

^ = to the power of
. = multiply
/ = divide

1. y = (x^2).(x^2) (I haven’t worked this one out yet but I can see an interesting proof coming along)
2. y = (x^3).(3x^2)
3. y = (2x).(e^x) (may be tricky for very new people, but we’ll get through it)
4. (y/2x^2) = x (a bit of re arranging to do)
5. y = ((2x + 3)^2)(3x^2) (tricky tricky)